Let $x_1,...,x_n $ are distinct real numbers.

Is it a formula for the Vandermonde type determinant $V(x_1, \cdots,x_n)$ whose last column is $x_1^k,\ \cdots,\ x_n^k$, where $k \geq n$, instead of $x_1^{n-1},\ \cdots,\ x_n^{n-1}$?


  • $\begingroup$ All the other columns are the same as usual? $\endgroup$ – EuYu Nov 16 '12 at 15:01
  • $\begingroup$ Yes, the first $n-1$ column are the same. $\endgroup$ – Richard Nov 16 '12 at 15:01

Sure, at least you can find such a formula for any fixed $k \geqslant n$. Not sure about a general formula for an unknown $k$ though.

Here is a hint: if all the $x_i$ are distinct, vector $(x_1^k,\ldots,x_n^k)$ is a linear combination of vectors $(x_1^i,\ldots,x_n^i)$, $i=0,\,1,\ldots,n-1$. If $$ (x_1^k,\ldots,x_n^k) = \sum_{i=0}^{n-1} \lambda_i (x_1^i,\ldots,x_n^i), $$ then your determinant is simply equal to $\lambda_{n-1}$ times the usual Vandermonde determinant. So all you need to do is find $\lambda_{n-1}$.


To continue Dan's answer, we want $$ x_i^k = \sum_{j=0}^{n-1} \lambda_j x_i^j, \qquad 1 \le i \le n $$ That is the polynomial $p(x) := \sum_{j=0}^{n-1} \lambda_j x^j$ interpolates $x^k$ at $x_0, \ldots, x_{n-1}$. Lagrange interpolation gives $$ p(x) = \sum_{j=0}^{n-1} x_j^k \cdot \prod_{\ell \ne j} \frac{x-x_\ell}{x_j - x_\ell} $$ $\lambda_{n-1}$ is the coefficient of $x^{n-1}$, so it is $$ \lambda_{n-1} = \sum_{j=0}^{n-1} x_j^k \prod_{\ell\ne j} \frac 1{x_j - x_\ell}. $$


Perform Laplace expansion along the last column. As the deletion of the $\ell$-th row and the last column gives a $(n-1)\times(n-1)$ Vandermonde matrix (in the original flavour), we get $$ \sum_{\ell=1}^n (-1)^{\ell+n} x_\ell^k\prod_{i<j\,\textrm{ and }\,i,j\not=\ell}(x_j-x_i). $$


Actually there is a general formula but not an easy way to describe it.

Let me first "define" the polynomials $f_m(x_1,x_2,\ldots,x_k)$ through some examples.

  • $f_0(x_1,x_2,x_3)=1.$
  • $f_1(x_1,x_2,x_3)=x_1+x_2+x_3.$
  • $f_2(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3.$
  • $f_3(x_1,x_2,x_3)=x_1^3+x_2^3+x_3^3+x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2+x_1x_2x_3.$
  • $f_3(x_1,x_2,x_3,x_4)=x_1^3+x_1^2x_2+x_1^2x_3+x_1^2x_4+x_1x_2^2+x_1x_2x_3+x_1x_2x_4+x_1x_3^2+x_1x_3x_4+x_1x_4^2+x_2^3+x_2^2x_3+x_2^2x_4+x_2x_3^2+x_2x_3x_4+x_2x_4^2+x_3^3+x_3^2x_4+x_3x_4^2+x_4^3 .$
  • $\ldots$

Let $V(x_1,x_2,\ldots,x_n,k)$ denote the Vandermonde type determinant whose last column is $x_1^k,x_2^k,\ldots,x_n^k$ where $k\geq n-1$. So $V(x_1,x_2,\ldots,x_n,n-1)$ is the usual Vandermonde. Then $$V(x_1,x_2,\ldots,x_n,k)=V(x_1,x_2,\ldots,x_n,n-1)\cdot f_{k-n+1}(x_1,x_2,\ldots,x_n).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.